A) OC
B) OD
C) OA
D) OB
Correct Answer: C
Solution :
1 kg of ice at \[0{{\,}^{o}}C\]is converted into water at \[0{{\,}^{o}}C.\] Heat taken by ice \[=mL\] \[=1\times 336\text{ }=336\,k\text{ joule}\] Heat given to water to raise its temp from \[0{{\,}^{o}}C\]to \[T{{\,}^{o}}C.\,T{{\,}^{o}}C\]is temperature of mixture. \[336+mc\,\Delta T={{m}_{\omega }}.c\Delta T\] \[336\,kJ+1\times 4.2\,kJ\] \[(T-O)=1\times 4.2\times kJ(80-T)\] \[336\times {{10}^{3}}+4.2\times {{10}^{3}}(T)=4.2\times {{10}^{3}}(80-T)\] \[336\times 4.2T=336-4.2\,T\] \[T=0{{\,}^{o}}C\]
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